100x + x - 50x + 100 - 50x = 100 . tìm x
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YC
1
LD
12 tháng 7 2018
\(x=49\Leftrightarrow50=x+1\)
Tính A = x7 - 50x6 + 50x5 - 50x4 + 50x3 - 50x2 +50x +100 với x = 49
\(\Rightarrow A=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x+100\)
\(=x^7-\left(x^7+x^6\right)+\left(x^6+x^5\right)-\left(x^5+x^4\right)+...+\left(x^2+1\right)+100\)
\(=x+100=149\)
VT
2
2 tháng 5 2015
\(x-\left(\frac{50x}{100}+\frac{25x}{100}\right)=\frac{45}{4}\)
\(x-\left(\frac{50x+25x}{100}\right)=\frac{45}{4}\)
\(x-\frac{75x}{100}=\frac{45}{4}\)
\(x-x\times\frac{3}{4}=\frac{45}{4}\)
\(x\times\left(1-\frac{3}{4}\right)=\frac{45}{4}\)
\(x\times\frac{1}{4}=\frac{45}{4}\)
\(x=\frac{45}{4}\div\frac{1}{4}\)
\(x=45\)
2 tháng 5 2015
\(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=\frac{45}{4}\)
\(x-\left(\frac{100x}{200}+\frac{25x}{200}\right)=\frac{45}{4}\)
\(x-\left(\frac{100x+25x}{200}\right)=\frac{45}{4}\)
\(x-\frac{125x}{200}=\frac{45}{4}\)
\(x-x\frac{5}{8}=\frac{45}{4}\)
\(x\left(1-\frac{5}{8}\right)=\frac{45}{4}\)
\(x.\frac{3}{8}=\frac{45}{4}\)
\(x\)=\(\frac{45}{4}:\frac{3}{8}\)
\(x\)=\(30\)
HT
4
16 tháng 7 2023
\(C=16x^2-8x+2024\)
\(\Rightarrow C=16x^2-8x+1+2023\)
\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)
\(\Rightarrow Min\left(C\right)=2023\)
\(D=-25x^2+50x-2023\)
\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)
\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=1998\)
\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)
\(\Rightarrow Max\left(B\right)=200\)
\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)
\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)
\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)
\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)
\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)
\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)
\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)
\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)
\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)
\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)
\(\Rightarrow Max\left(F\right)=48\)
OY
15 tháng 8 2021
\(x-\left(\dfrac{50x}{100}+\dfrac{25x}{100}\right)=11\dfrac{1}{4}\)
\(x-\dfrac{3x}{4}=\dfrac{45}{4}\)
\(\dfrac{x}{4}=\dfrac{45}{4}\Rightarrow x=45\)
NQ
1
NT
2
TT
30 tháng 5 2018
a) \(2^{x+2}-2^2=96\)
<=> \(2^x.2^2-2^x=96\)
<=> \(2^x\left(4-1\right)=96\)
<=> \(3.2^x=96\)
<=> \(2^x=32\)
<=> \(2^x=2^5\)
<=> x = 5
31 tháng 5 2018
b, \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\)
\(\Rightarrow x-\left(\frac{1x}{2}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\left(\frac{4x}{8}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{8x}{8}-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\Rightarrow x=\frac{45}{4}\div\frac{3}{8}=30\)
Vậy x = 30
HP
0
( 100x - 50x - 50x ) + x + 100 = 100
=> x + 100 = 100
=> x = 0
Ta có: 100x + x - 50x + 100 - 50x = 100
=> (100x + x - 50 - 50x) + 100 = 100
=> x + 100 = 100
=> x = 100 - 100
=> x = 0